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Proof step 25 from proof subarray-eq-elim (previous step)

# ar = m + n, ar = subarray(ar, 0, n) + subarray(ar, n, m) ⊦ subarray(ar, n, m) = ar1 ↔ (∃ ar0, ar2. ar = ar0 + ar1 + ar2 ∧ # ar0 = n ∧ # ar2 = # ar1 + # ar2 - m)

User applied rule case distinction

And obtained the following new goals

Proof step 26:
# ar = m + n, ar = subarray(ar, 0, n) + subarray(ar, n, m), subarray(ar, n, m) = ar1 ⊦ ∃ ar0, ar2. ar = ar0 + ar1 + ar2 ∧ # ar0 = n ∧ # ar2 = # ar1 + # ar2 - m
Proof step 30:
n + m ≤ # ar0 ⊦ subarray(ar0 + ar1, n, m) = subarray(ar0, n, m)
Proof step 31:
⊦ (ar + ar0) + ar1 = ar + ar0 + ar1
Proof step 32:
⊦ ar + ar0 = ar + ar1 ↔ ar0 = ar1
Proof step 33:
⊦ #(ar0 + ar1) = # ar0 + # ar1
Proof step 34:
⊦ ar0 + ar1 = ar0 ↔ # ar1 = 0

Rule argument

pos: right 1

Additional information

Used simplifier rules: subarray-elim-right: n + m ≤ # ar0 ⊦ subarray(ar0 + ar1, n, m) = subarray(ar0, n, m);%append-a: ⊦ (ar + ar0) + ar1 = ar + ar0 + ar1;%append-eq-left: ⊦ ar + ar0 = ar + ar1 ↔ ar0 = ar1;%append-len: ⊦ #(ar0 + ar1) = # ar0 + # ar1;%append-prefix-left: ⊦ ar0 + ar1 = ar0 ↔ # ar1 = 0;% ⊦ n + 0 = n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ ¬ n < n; (s)%# ar = n ⊦ subarray(ar, 0, n) = ar; (s)% ⊦ m + n = m + n0 ↔ n = n0; (s)% ⊦ m + m0 ≤ m ↔ m0 = 0; (s)% ⊦ m + n - n + n1 = m - n1; (s)% ⊦ m - m = 0; (s)% ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; (ws)